如何在排序后减去天数 [英] How to Subtract Days after Sorting

问题描述

按照人们的生日排序后，我如何减去两个生日以找出两者之间的天数，以便知道某人左边或右边的人的生日最接近? br/>
例如:
1月31日---- 2月2日---- 2月15日
Person1 --- Person2 --- Person3

结果:Person2的生日最接近person1.

如果您想知道the年，则除非某人有b天，否则2月仅算作28天.

我应该将1月= 1，就像我应该放
#define 1月1日，然后每月执行一次?

那我可以做些什么:

if (month == JANUARY)
  dayofyear = day;
else if (month == MARCH)
  dayofyear = day+28+31;
...and so on

*仅C编程.

谢谢，我终于想到了这样的东西:

int dayOfYear = dayOfMonth + monthOffset[monthNumber];

但是没有意识到大约12月31日和1月1日.人们是按照生日而不是年龄来排序的，所以问题在于忽略了这一年，所以我猜想这一年没有发挥作用.我想我只需要拿出另一个函数或if语句来表示12月

解决方案

您的麻烦是什么?
>

您可以基于一年中的某天(可以通过简单的表格查询来计算):

int dayOfYear = dayOfMonth + monthOffset[monthNumber];

其中monthOffset是一个数组，其中包含12个常数，从0和31开始.您可能有两个常数，一个代表leap年，一个正则.

但是，新年前后您可能会遇到问题.如果时间跨度超过半年，则应将其镜像到下一年.

:)

After people are sorted in order of their birthdays, how can I subtract two birthdays to find out the days in between so that I can know if the person to the left or the right of someone has the closest birthday to them?

example:
JAN 31 ---- FEB 2 ---- FEB 15
Person1 --- Person2 --- Person3

Result: Person2 has closest birthday to person1.

In case you wonder about the leap year, February is only be counted as 28 days, unless someone has that b-day.

Should I make January = 1, like should I put
#define JANUARY 1 and do that for each month?

Then could I do something like:

if (month == JANUARY)
  dayofyear = day;
else if (month == MARCH)
  dayofyear = day+28+31;
...and so on

*C programming only.

Thanks, yeah I finally thought of something like that:

int dayOfYear = dayOfMonth + monthOffset[monthNumber];

but didn''t realize about December 31 and Jan 1. People are being sorted by their birthdays, not age, so the problem says to ignore the year for that so I guess the year doesn''t come into play. I guess I will just have to come up with another function or if statement for December

解决方案

What is your trouble about?
A little more than a switch will do the job.

Hi,

you can base this on day-of-year, which can be calculated as a simple table look-up:

int dayOfYear = dayOfMonth + monthOffset[monthNumber];

where monthOffset is an array holding 12 constants, starting with 0 and 31. You may have two of them, one for leap years, one regular.

However, you have a potential problem around New Year. When a time span exceeds half a year, you should mirror it to the next year.

:)

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